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Sagot :
Certainly! Let's solve each part of the question step-by-step:
### Part (a)
We need to find the order (dimensions) of matrix [tex]\( M \)[/tex] and its transpose [tex]\( M^\top \)[/tex].
Matrix [tex]\( M \)[/tex]:
[tex]\[ M = \begin{pmatrix} 1 & 5 \\ 2 & 7 \end{pmatrix} \][/tex]
The matrix [tex]\( M \)[/tex] has 2 rows and 2 columns, so the order of [tex]\( M \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
Transpose of Matrix [tex]\( M \)[/tex]:
[tex]\[ M^\top = \begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix} \][/tex]
The transpose [tex]\( M^\top \)[/tex] also has 2 rows and 2 columns, so the order of [tex]\( M^\top \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
Conclusion:
The order of [tex]\( M \)[/tex] and [tex]\( M^\top \)[/tex] is the same, both are [tex]\( 2 \times 2 \)[/tex].
### Part (b)
We need to find the order (dimensions) of matrix [tex]\( N \)[/tex] and its transpose [tex]\( N^\top \)[/tex].
Matrix [tex]\( N \)[/tex]:
[tex]\[ N = \begin{pmatrix} -3 & 6 \\ 2 & 8 \end{pmatrix} \][/tex]
The matrix [tex]\( N \)[/tex] has 2 rows and 2 columns, so the order of [tex]\( N \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
Transpose of Matrix [tex]\( N \)[/tex]:
[tex]\[ N^\top = \begin{pmatrix} -3 & 2 \\ 6 & 8 \end{pmatrix} \][/tex]
The transpose [tex]\( N^\top \)[/tex] also has 2 rows and 2 columns, so the order of [tex]\( N^\top \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
Conclusion:
The order of [tex]\( N \)[/tex] and [tex]\( N^\top \)[/tex] is the same, both are [tex]\( 2 \times 2 \)[/tex].
### Part (c)
We need to prove that the double transpose of [tex]\( M \)[/tex] equals [tex]\( M \)[/tex]:
[tex]\[ (M^\top)^\top = M \][/tex]
Step-by-Step Verification:
1. Calculate [tex]\( M^\top \)[/tex]:
[tex]\[ M^\top = \begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix} \][/tex]
2. Calculate the transpose of [tex]\( M^\top \)[/tex]:
[tex]\[ (M^\top)^\top = \begin{pmatrix} 1 & 5 \\ 2 & 7 \end{pmatrix} \][/tex]
Conclusion:
[tex]\( (M^\top)^\top = M \)[/tex] is true.
### Part (d)
We need to prove that the double transpose of [tex]\( N \)[/tex] equals [tex]\( N \)[/tex]:
[tex]\[ (N^\top)^\top = N \][/tex]
Step-by-Step Verification:
1. Calculate [tex]\( N^\top \)[/tex]:
[tex]\[ N^\top = \begin{pmatrix} -3 & 2 \\ 6 & 8 \end{pmatrix} \][/tex]
2. Calculate the transpose of [tex]\( N^\top \)[/tex]:
[tex]\[ (N^\top)^\top = \begin{pmatrix} -3 & 6 \\ 2 & 8 \end{pmatrix} \][/tex]
Conclusion:
[tex]\( (N^\top)^\top = N \)[/tex] is true.
### Part (e)
We need to prove that the transpose of the sum of matrices [tex]\( M \)[/tex] and [tex]\( N \)[/tex] equals the sum of their transposes:
[tex]\[ (M + N)^\top = M^\top + N^\top \][/tex]
Step-by-Step Verification:
1. Calculate [tex]\( M + N \)[/tex]:
[tex]\[ M + N = \begin{pmatrix} 1 & 5 \\ 2 & 7 \end{pmatrix} + \begin{pmatrix} -3 & 6 \\ 2 & 8 \end{pmatrix} = \begin{pmatrix} -2 & 11 \\ 4 & 15 \end{pmatrix} \][/tex]
2. Calculate the transpose of [tex]\( M + N \)[/tex]:
[tex]\[ (M + N)^\top = \begin{pmatrix} -2 & 4 \\ 11 & 15 \end{pmatrix} \][/tex]
3. Calculate [tex]\( M^\top \)[/tex]:
[tex]\[ M^\top = \begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix} \][/tex]
4. Calculate [tex]\( N^\top \)[/tex]:
[tex]\[ N^\top = \begin{pmatrix} -3 & 2 \\ 6 & 8 \end{pmatrix} \][/tex]
5. Calculate the sum of [tex]\( M^\top \)[/tex] and [tex]\( N^\top \)[/tex]:
[tex]\[ M^\top + N^\top = \begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix} + \begin{pmatrix} -3 & 2 \\ 6 & 8 \end{pmatrix} = \begin{pmatrix} -2 & 4 \\ 11 & 15 \end{pmatrix} \][/tex]
Conclusion:
[tex]\( (M + N)^\top = M^\top + N^\top \)[/tex] is true.
In summary:
a) The order of [tex]\( M \)[/tex] and [tex]\( M^\top \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
b) The order of [tex]\( N \)[/tex] and [tex]\( N^\top \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
c) [tex]\( (M^\top)^\top = M \)[/tex].
d) [tex]\( (N^\top)^\top = N \)[/tex].
e) [tex]\( (M + N)^\top = M^\top + N^\top \)[/tex].
### Part (a)
We need to find the order (dimensions) of matrix [tex]\( M \)[/tex] and its transpose [tex]\( M^\top \)[/tex].
Matrix [tex]\( M \)[/tex]:
[tex]\[ M = \begin{pmatrix} 1 & 5 \\ 2 & 7 \end{pmatrix} \][/tex]
The matrix [tex]\( M \)[/tex] has 2 rows and 2 columns, so the order of [tex]\( M \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
Transpose of Matrix [tex]\( M \)[/tex]:
[tex]\[ M^\top = \begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix} \][/tex]
The transpose [tex]\( M^\top \)[/tex] also has 2 rows and 2 columns, so the order of [tex]\( M^\top \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
Conclusion:
The order of [tex]\( M \)[/tex] and [tex]\( M^\top \)[/tex] is the same, both are [tex]\( 2 \times 2 \)[/tex].
### Part (b)
We need to find the order (dimensions) of matrix [tex]\( N \)[/tex] and its transpose [tex]\( N^\top \)[/tex].
Matrix [tex]\( N \)[/tex]:
[tex]\[ N = \begin{pmatrix} -3 & 6 \\ 2 & 8 \end{pmatrix} \][/tex]
The matrix [tex]\( N \)[/tex] has 2 rows and 2 columns, so the order of [tex]\( N \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
Transpose of Matrix [tex]\( N \)[/tex]:
[tex]\[ N^\top = \begin{pmatrix} -3 & 2 \\ 6 & 8 \end{pmatrix} \][/tex]
The transpose [tex]\( N^\top \)[/tex] also has 2 rows and 2 columns, so the order of [tex]\( N^\top \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
Conclusion:
The order of [tex]\( N \)[/tex] and [tex]\( N^\top \)[/tex] is the same, both are [tex]\( 2 \times 2 \)[/tex].
### Part (c)
We need to prove that the double transpose of [tex]\( M \)[/tex] equals [tex]\( M \)[/tex]:
[tex]\[ (M^\top)^\top = M \][/tex]
Step-by-Step Verification:
1. Calculate [tex]\( M^\top \)[/tex]:
[tex]\[ M^\top = \begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix} \][/tex]
2. Calculate the transpose of [tex]\( M^\top \)[/tex]:
[tex]\[ (M^\top)^\top = \begin{pmatrix} 1 & 5 \\ 2 & 7 \end{pmatrix} \][/tex]
Conclusion:
[tex]\( (M^\top)^\top = M \)[/tex] is true.
### Part (d)
We need to prove that the double transpose of [tex]\( N \)[/tex] equals [tex]\( N \)[/tex]:
[tex]\[ (N^\top)^\top = N \][/tex]
Step-by-Step Verification:
1. Calculate [tex]\( N^\top \)[/tex]:
[tex]\[ N^\top = \begin{pmatrix} -3 & 2 \\ 6 & 8 \end{pmatrix} \][/tex]
2. Calculate the transpose of [tex]\( N^\top \)[/tex]:
[tex]\[ (N^\top)^\top = \begin{pmatrix} -3 & 6 \\ 2 & 8 \end{pmatrix} \][/tex]
Conclusion:
[tex]\( (N^\top)^\top = N \)[/tex] is true.
### Part (e)
We need to prove that the transpose of the sum of matrices [tex]\( M \)[/tex] and [tex]\( N \)[/tex] equals the sum of their transposes:
[tex]\[ (M + N)^\top = M^\top + N^\top \][/tex]
Step-by-Step Verification:
1. Calculate [tex]\( M + N \)[/tex]:
[tex]\[ M + N = \begin{pmatrix} 1 & 5 \\ 2 & 7 \end{pmatrix} + \begin{pmatrix} -3 & 6 \\ 2 & 8 \end{pmatrix} = \begin{pmatrix} -2 & 11 \\ 4 & 15 \end{pmatrix} \][/tex]
2. Calculate the transpose of [tex]\( M + N \)[/tex]:
[tex]\[ (M + N)^\top = \begin{pmatrix} -2 & 4 \\ 11 & 15 \end{pmatrix} \][/tex]
3. Calculate [tex]\( M^\top \)[/tex]:
[tex]\[ M^\top = \begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix} \][/tex]
4. Calculate [tex]\( N^\top \)[/tex]:
[tex]\[ N^\top = \begin{pmatrix} -3 & 2 \\ 6 & 8 \end{pmatrix} \][/tex]
5. Calculate the sum of [tex]\( M^\top \)[/tex] and [tex]\( N^\top \)[/tex]:
[tex]\[ M^\top + N^\top = \begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix} + \begin{pmatrix} -3 & 2 \\ 6 & 8 \end{pmatrix} = \begin{pmatrix} -2 & 4 \\ 11 & 15 \end{pmatrix} \][/tex]
Conclusion:
[tex]\( (M + N)^\top = M^\top + N^\top \)[/tex] is true.
In summary:
a) The order of [tex]\( M \)[/tex] and [tex]\( M^\top \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
b) The order of [tex]\( N \)[/tex] and [tex]\( N^\top \)[/tex] is [tex]\( 2 \times 2 \)[/tex].
c) [tex]\( (M^\top)^\top = M \)[/tex].
d) [tex]\( (N^\top)^\top = N \)[/tex].
e) [tex]\( (M + N)^\top = M^\top + N^\top \)[/tex].
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